Dummit And Foote Solutions Chapter 4 Overleaf High Quality -

\beginsolution Let $|H| = n$ and suppose $H$ is the only subgroup of $G$ with order $n$. For any $g \in G$, consider $gHg^-1$. Conjugation is an automorphism of $G$, so $|gHg^-1| = |H| = n$. Thus $gHg^-1$ is also a subgroup of $G$ of order $n$. By uniqueness, $gHg^-1 = H$ for all $g \in G$. Hence $H \trianglelefteq G$. \endsolution

% Theorem-like environments \newtheorem*propositionProposition \newtheorem*lemmaLemma Dummit And Foote Solutions Chapter 4 Overleaf High Quality

\subsection*Exercise 4.3.12 \textitProve that if $H$ is the unique subgroup of a finite group $G$ of order $n$, then $H$ is normal in $G$. \beginsolution Let $|H| = n$ and suppose $H$

\subsection*Exercise 4.2.6 \textitLet $G$ be a group and let $H$ be a subgroup of $G$. Prove that $C_G(H) \le N_G(H)$. r^2\ \cong \Z/2\Z$. \endsolution

Hence $Z(D_8) = \1, r^2\ \cong \Z/2\Z$. \endsolution